3a) Note that the problem says that the tickets are bought at the same time for the same price. This makes the formula 3C + 5F.
If the each ticket was bought separately, that would make the formula C + C + C + F + F + ... The mean is the same for either formula. But if the second formula was used, you'd have to do the Pythagorean formula thing.
3b) First do 3C = 450 and 5F = 500; then do sqrt(450^2 + 500^2)
3c) This is only about ONE of each tix: C - F. You can take it from there...
6) Make a 2-way table: 51% in the upper left corner. The other two numbers go on the outside.
a) only 3% is left in the lower right corner
b) P(left|right) = 51/82 = 62.2%. Since P(left) = 66%, this is fairly different, so they are not independent.
7a) 1 - (89/90)^10
7b) 1 - (9/10)^10
7c) 1 - (89/90)^5*(9/10)^5
25d) (0.93)^4*(0.07)
28a) mean = 4
28b) sd = 3.2
28c) Think and read carefully! If the first is bigger than the second then:
(first - second) > zero
So we want to use the mean and sd from above to compare to zero:
z = (0 - 4)/3.2 = -1.249
P(z>-1.249) = (using normalcdf) = 89.4%
42a) 1/100 = 0.01
42b) (.99)(.99)(.01) = 0.009801
42c) (.99)^100 = 0.366
42d) You want to be first!
42e) It doesn't matter! Everyone has a 1% chance!
Now if you're thinking carefully about (e), you might be thinking: "Hey, don't the probabilities change?" Watch this!
Prob(3rd person wins) = (99/100)*(98/99)*(1/98) = 1% (notice how all the fractions reduce!)
Pretty cool, huh?
I will check e-mail at about 10-ish tonight. If you are feeling frustrated, drop a line to:
mrmathman @ gmail.com
I will reply tonight.
Good luck!
